Lemma 36.2.3. Any thickening of an affine scheme is affine.

**Proof.**
This is a special case of Limits, Proposition 31.11.2.
$\square$

**Proof for a finite order thickening.**
Suppose that $X \subset X'$ is a finite order thickening with $X$ affine. Then we may use Serre's criterion to prove $X'$ is affine. More precisely, we will use Cohomology of Schemes, Lemma 29.3.1. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_{X'}$-module. It suffices to show that $H^1(X', \mathcal{F}) = 0$. Denote $i : X \to X'$ the given closed immersion and denote $\mathcal{I} = \mathop{\mathrm{Ker}}(i^\sharp : \mathcal{O}_{X'} \to i_*\mathcal{O}_ X)$. By our discussion of finite order thickenings (following Definition 36.2.1) there exists an $n \geq 0$ and a filtration

by quasi-coherent submodules such that $\mathcal{F}_ a/\mathcal{F}_{a + 1}$ is annihilated by $\mathcal{I}$. Namely, we can take $\mathcal{F}_ a = \mathcal{I}^ a\mathcal{F}$. Then $\mathcal{F}_ a/\mathcal{F}_{a + 1} = i_*\mathcal{G}_ a$ for some quasi-coherent $\mathcal{O}_ X$-module $\mathcal{G}_ a$, see Morphisms, Lemma 28.4.1. We obtain

The second equality comes from Cohomology of Schemes, Lemma 29.2.4 and the last equality from Cohomology of Schemes, Lemma 29.2.2. Thus $\mathcal{F}$ has a finite filtration whose successive quotients have vanishing first cohomology and it follows by a simple induction argument that $H^1(X', \mathcal{F}) = 0$. $\square$

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